Padé Approximation and Apostol-Bernoulli and -Euler Polynomials
Marc Prévost
1
Univ Lille Nord de France, F-59000 Lille, France
2
ULCO, LMPA J. Liouville, B.P. 699, F-62228 Calais
3
CNRS, FR 2956
1,2,3
Abstract
Using the Padé approximation of the exponential function, we obtain recurrence relations between
Apostol-Bernoulli and between Apostol-Euler polynomials. As applications, we derive some new
lacunary recurrence relations for Bernoulli and Euler polynomials with gap of length 4 and lacunary
relations for Bernoulli and Euler numbers with gap of length 6.
Key words: Apostol-Bernoulli polynomials, Apostol-Euler polynomials, Bernoulli numbers,
Euler numbers, Padé approximants
2000 MSC: 11B68, 41A21
1. Introduction
Classical Bernoulli and Euler polynomials play a fundamental role in various branches of
mathematics including combinatorics, number theory and special functions. They are usually
defined by means of the following generating functions
t
ex t
t
e −1
=
2
ex t
+1
=
et
∞
X
k=0
∞
X
Bk (x)
tk
,
k!
(|t| < 2π),
(1.1)
Ek (x)
tk
,
k!
(|t| < π).
(1.2)
k=0
The first polynomials are
B0 (x)
=
1,
B1 (x)
= x − 1/2,
B2 (x)
2
= x − x + 1/6,
E0 (x) = 1,
E1 (x) = x − 1/2,
E2 (x) = x2 − x.
From (1.1) and (1.2), it is easily proved that
Bn (1 − x) = (−1)n Bn (x),
E n (1 − x) = (−1)n En (x),
Bn (x + 1) − Bn (x) = n xn−1 ,
n X
n
Bn (x + y) =
Bk (x)y n−k ,
k
E n (x + 1) + En (x) = 2 xn ,
n X
n
E n (x + y) =
Ek (x)y n−k .
k
k=0
Email address: [email protected] (Marc Prévost
Preprint submitted to Elsevier
k=0
1,2,3 )
November 24, 2009
The classical Bernoulli numbers Bn and the classical Euler numbers En are given by Bn := Bn (0)
and En := 2n En (1/2), respectively.
In 2000, T. Agoh [1] proved a general linear recurrence relation between Bernoulli and Euler
polynomials.
More recently, in [2], Chen and Sun made use of Zeilberger’s algorithm to prove most of the
existing recurrence relations for Bernoulli and Euler polynomials. They also derived two new
identities which are particular cases of the main theorem of this paper.
Some analogues of the classical Bernoulli polynomials were introduced by Apostol in order to
evaluate the Hurwitz-Lerch zeta function:
φ(z, s, a) :=
∞
X
zn
.
(a + n)s
n=0
See [3] and also the recent book [4].
We begin by recalling here Apostol’s definition as follows:
Definition 1. (Apostol, [3]) The Apostol-Bernoulli polynomials Bk (x; λ) in the variable x are
defined by means of the following generating function:
t
ex t
λ et − 1
∞
X
=
Bk (x; λ)
k=0
tk
k!
(1.3)
(|t| < 2π when λ = 1; |t| < |log λ| when λ 6= 1 ).
Using the Gaussian hypergeometric functions, Luo [5] found formulas for Bk (x; λ) and in [6],
Boyadzhiev found the relations of the Apostol-Bernoulli functions with the Euler polynomials and
the derivative polynomials for the cotangent function.
Recently, Luo and Srivastava introduced the Apostol-Bernoulli polynomials of higher order
(also called generalized Apostol-Bernoulli polynomials):
(α)
Definition 2. (Luo and Srivastava, [7]) The Apostol-Bernoulli polynomials Bk (x; λ) of order α
in the variable x are defined by means of the generating function:
t
λ et − 1
α
xt
e
=
∞
X
(α)
Bk (x; λ)
k=0
tk
,
k!
(1.4)
(|t| < 2π when λ = 1; |t| < |log λ| when λ 6= 1).
According to the definition, by setting α = 1, we obtain the Apostol-Bernoulli polynomials
Bk (x; λ). Moreover, we call Bk (λ) := Bk (0; λ) the Apostol-Bernoulli numbers.
(α)
Explicit representation of Bk (x; λ) in terms of a generalization of the Hurwitz-Lerch zeta
function can be found in [8].
In the paper [9] submitted in 2004, Luo introduced Apostol-Euler polynomials of higher order
α:
(α)
Definition 3. (Luo, [9]) The Apostol-Euler polynomials Ek (x; λ) of order α in the variable x
are defined by means of the following generating function:
2
λ et + 1
α
ex t
=
∞
X
(α)
Ek (x; λ)
k=0
2
tk
, (λ 6= −1, |t| < |log(−λ)|).
k!
(1.5)
(1)
The Apostol-Euler polynomials Ek (x; λ) are given by Ek (x; λ) := Ek (x; λ). The Apostol-Euler
numbers Ek (λ) are given by Ek (λ) := 2k Ek ( 21 ; λ).
Some relations between Apostol-Bernoulli and Apostol-Euler polynomials of order α can be
found in [10]. For more results on these polynomials, the readers are referred to [11, 12, 13].
In this paper, we only consider Apostol-type polynomials Bk (x; λ) and Ek (x; λ). For λ = 1,
they reduce to the classical Bernoulli polynomials Bk (x) and the classical Euler polynomials Ek (x),
respectively. For λ 6= 1, deg Bk (x; λ) = k − 1. For λ 6= −1, deg Ek (x; λ) = k. The first expressions
of these polynomials are
B0 (x; λ) = 0,
2
,
1+λ
2(x − λ + xλ)
,
E1 (x; λ) =
(1 + λ)2
2(x2 − λ − 2xλ + 2x2 λ + λ2 − 2xλ2 + x2 λ2 )
E2 (x; λ) =
.
(1 + λ)3
E0 (x; λ) =
1
,
λ−1
2(−x − λ + xλ)
B2 (x; λ) =
,
(λ − 1)2
B1 (x; λ) =
The core of the paper is Theorem 1. Using Padé approximation to et in the generating functions
(1.3) and (1.5), we get a new relation between Apostol-type polynomials depending on three
parameters n, m, p where n and m are respectively the degree of the numerator and the degree
of the denominator of the Padé approximant used to approximate the function et and p is some
positive integer. We will show that many known recurrence relations are particular cases of this
general formula.
The paper is organized as follows. In the next section, we recall the definition of Padé approximant to a general series and its expression for the case of the exponential function. In Section 3,
we apply Padé approximation to prove the announced general recurrence relations (Theorem 1)
for the Apostol-Bernoulli and Apostol-Euler polynomials. In Section 4 we show that many known
linear recurrence relations are particular cases of the general recurrence relations. A generalization
of Theorem 1 is given in Section 5. In the last two sections, as applications, we will give some
lacunary recurrence relations with a gap of length 4 between Bernoulli and Euler polynomials and
with a gap of length 6 between Bernoulli and Euler numbers.
2. Padé approximant
In this section, we recall the definition of Padé approximation to general series and their
expression in the case of the exponential function. Given a function f with a Taylor expansion
f (t) =
∞
X
ci ti
(2.1)
i=0
in a neighborhood of the origin, a Padé approximant denoted [n, m]f to f is a rational fraction of
degree n (resp. m) for the numerator (resp. the denominator):
[n, m]f (t) =
α0 + α1 t + · · · + αn tn
,
β0 + β1 t + · · · + βm tm
whose Taylor expansion agrees with (2.1) as far as possible:
∞
X
i=0
ci t i −
α0 + α1 t + · · · + αn tn
= O(tm+n+1 ).
β0 + β1 t + · · · + βm tm
In the general case, the resulting linear system has unique solutions αi , βi (see, e.g., [14]).
Padé approximation is related with convergence acceleration [15], continued fractions [16, 17],
orthogonal polynomials and quadrature formulas [18]. Moreover the denominators of Padé approximants satisfy a three terms recurrence [19, 20] and this property allows finding another proof
of the irrationality of ζ(2) and ζ(3) (see [21]).
3
If f (t) = et then
P (n,m) (t)
1 F1 (−n, −m − n, t)
=
Q(n,m) (t)
1 F1 (−m, −m − n, −t)
Pn
n
j
Pn
(−n)j tj
j=0 j (n + m − j)!t
j=0 (−m−n)j j!
= Pm
=
,
(−m)j (−t)j
Pm
m
j
j=0 (−m−n)j j!
(n
+
m
−
j)!(−t)
j=0
j
[n, m]f (t) : =
where the Pochhammer symbol (a)j is defined as
(a)j = a(a + 1) · · · (a + j − 1) if j ≥ 1,
= 1 if j = 0
P∞
k
k z
and the hypergeometric series 1 F1 (a, b, z) is defined as 1 F1 (a, b, z) := k=0 (a)
(b)k k! . In the sequel
we write [n, m](t) for the Padé approximant to et . The remainder term is defined by
R(n,m) (t) := et − [n, m](t) = et −
P (n,m) (t)
Q(n,m) (t)
and satisfies
et
R(n,m) (t) = tm+n+1
=
Q(n,m) (t)
∞
(−1)m X
Q(n,m) (t)
j=0
1
Z
(x − 1)m xn e−x t dx
0
tj+m+n+1
m+n+j
(j + m + n + 1)
j!
n
∞
=
(m + j)!
(−1)m n! X
tj+m+n+1
Q(n,m) (t) j=0 j! (m + n + j + 1)!
= O(tn+m+1 ).
In the sequel, let
(n,m)
αj
(n,m)
βj
(n,m)
γj
n
:=
(n + m − j)!, 0 ≤ j ≤ n,
j
m
:=
(n + m − j)!(−1)j , 0 ≤ j ≤ m,
j
:= (−1)m
n! (m + j)!
,
(m + n + 1 + j)! j!
j ≥ 0.
The polynomials P (n,m) , Q(n,m) and the product R(n,m) Q(n,m) are then given by
P (n,m) (t) =
Q(n,m) (t) =
R(n,m) (t) Q(n,m) (t) =
n
X
j=0
m
X
j=0
∞
X
j=0
4
(n,m) j
αj
t ,
(n,m) j
βj
t ,
(n,m) j+m+n+1
γj
t
.
3. Recurrence relations for Apostol-type polynomials
Let us recall one method to derive the basic formula for Bernoulli numbers. It consists in
considering the Taylor expansion of et around t = 0.
The generating function of the Bernoulli numbers Bk := Bk (0; 1) can be written as
t
(e − 1)
∞
X
k=0
∞
∞ j X
X
tk
tk
t
Bk = t,
Bk = t ⇐⇒
k!
j!
k!
j=1
(|t| < 2π).
k=0
So, using the Cauchy product for series,
∞
X
X
tp
p=1
k+j=p
k≥0,j≥1
Bk
= t.
k! j!
Pp−1 This implies B0 = 1 and j=0 pj Bj = 0 for p ≥ 2.
In this section, the main idea is to replace in the generating functions of Apostol-type polynomials (1.3) and (1.5), the exponential function et not by its Taylor expansion around t = 0 but by
its Padé approximant [n, m](t). This gives the following main result.
Theorem 1. For all integers m ≥ 0, n ≥ 0, the linear combination of Apostol-Bernoulli polynomials defined by
max(n,m) (n + m − j)!
n
m
− (−1)j
Bp−j (x; λ)
j
j
(p − j)!
j=0
n m
X
X
n (n + m − j)!
m (n + m − j)!
=λ
Bp−j (x; λ) −
(−1)j
Bp−j (x; λ)
j
j
(p − j)!
(p − j)!
j=0
j=0
A(p)
n,m (x; λ) :=
X
λ
satisfies for 0 ≤ p ≤ m + n,
A(p)
n,m (x; λ) =
p−1 X
m
j=0
j
(n + m − j)!(−1)j
xp−1−j
,
(p − 1 − j)!
(3.1)
and for p≥ m + n + 1,
A(p)
n,m (x; λ) =
m X
m
j=0
j
(n+m − j)!(−1)j
λ (−1)
m n!m!
xp−1−j
−
(p − 1 − j)!
p−m−n−1
X p!
k=0
p−1−n−k
m
p
Bk (x; λ).
k
(3.2)
We use the convention Bk (x; λ) = 0 for k ≤ −1.
Also, for all integers m ≥ 0, n ≥ 0, the linear combination of Apostol-Euler polynomials defined
by
max(n,m) (n + m − j)!
n
m
+ (−1)j
Ep−j (x; λ)
j
j
(p − j)!
j=0
n m
X
X
n (n + m − j)!
j m (n + m − j)!
=λ
Ep−j (x; λ) +
(−1)
Ep−j (x; λ)
j
j
(p − j)!
(p − j)!
j=0
j=0
b(p)
A
n,m (x; λ) :=
X
λ
satisfies for 0 ≤ p ≤ m + n,
5
p X
m
b(p)
A
n,m (x; λ) = 2
(n + m − j)!(−1)j
j
j=0
xp−j
,
(p − j)!
(3.3)
and for p ≥ m + n + 1,
b(p)
A
n,m (x; λ)
=
2
m X
m
j
j=0
(n + m − j)!(−1)j
n! m!
p!
m
λ (−1)
p−m−n−1
X k=0
xp−j
−
(p − j)!
p−1−n−k
m
p
E (x; λ).
k k
(3.4)
We use the convention Ek (x; λ) = 0 for k ≤ −1.
Proof. We start from the generating function (1.3),
∞
X
tk
= t ex t ,
Bk (x; λ)
λ et − 1
k!
(λ = 1, |t| < 2π; λ 6= 1, |t| < |log λ|),
k=0
and for some integers n, m, we replace et by its Padé approximant, previously defined,
P (n,m) (t)
+ R(n,m) (t).
Q(n,m) (t)
et = [n, m](t) + R(n,m) (t) =
We get
X
(n,m)
∞
tk
P
(t)
Bk (x; λ) = t ex t
+ λR(n,m) (t) − 1
λ (n,m)
k!
Q
(t)
k=0
and
∞
X
tk
Bk (x; λ) = t ex t Q(n,m) (t).
λP (n,m) (t) − Q(n,m) (t) + λR(n,m) (t) Q(n,m) (t)
k!
k=0
Thus
λP (n,m) (t) − Q(n,m) (t)
∞
∞
X
Bk (x; λ)
k=0
X
tk
tk
Bk (x; λ) .
= t ex t Q(n,m) (t) − λR(n,m) (t) Q(n,m) (t)
k!
k!
k=0
From the expression of the Padé approximant, it arises the following relation
max(n,m) X
(n,m)
λαj
−
(n,m)
βj
t
j=0
j
∞
X
Bk (x; λ)
k=0
∞
m
X
xk t k X
t
k!
k=0
tk
=
k!
(n,m) j
t −λ
βj
j=0
∞
X
(n,m) j+m+n+1
γj
t
j=0
∞
X
Bk (x; λ)
k=0
tk
k!
which gives, applying the Cauchy product for series,
∞
X
p=0
tp
X
(n,m)
λαj
(n,m)
− βj
k+j=p
k≥0, j≥0
∞
X
p=0
tp
B (x; λ)
k
=
k!
X
k+j=p−1
k≥0, j≥0
(n,m) x
βj
k
k!
6
−λ
∞
X
p=0
tp
X
k+j=p−m−n−1
k≥0, j≥0
(n,m) Bk (x; λ)
γj
k!
.
Equating the coefficients of tp gives for 0 ≤ p ≤ m + n
X
(n,m)
λαj
(n,m)
− βj
k+j=p
k≥0, j≥0
B (x; λ)
k
=
k!
(n,m) x
X
βj
k
,
k!
k+j=p−1
k≥0, j≥0
and for p = r + m + n, r ≥ 1
X
(n,m)
λαj
(n,m)
− βj
k+j=p
k≥0, j≥0
B (x; λ)
k
=
k!
(n,m) x
X
βj
k
k!
k+j=p−1
k≥0, j≥0
(n,m) Bk (x; λ)
X
−λ
γj
k!
k+j=r−1
k≥0, j≥0
,
(p)
which proves the expression of An,m (x; λ) in Theorem 1.
b(p)
For A
n,m (x; λ) the proof is similar. We start from the generating function (1.5)
λ et + 1
∞
X
Ek (x; λ)
k=0
tk
= 2 ex t ,
k!
(|t| < |log(−λ)|),
and as previously we replace et by its Padé approximant [n, m](t). We get
∞
X
p=0
X
tp
(n,m)
λαj
(n,m)
+ βj
k+j=p
k≥0, j≥0
E (x; λ)
k
k!
=
2
∞
X
p=0
λ
(n,m) x
X
tp
βj
k!
k+j=p
k≥0, j≥0
∞
X
p=0
−
(n,m) Ek (x; λ)
X
tp
k
γj
k+j=p−m−n−1
k≥0, j≥0
k!
.
So, for 0 ≤ p ≤ m + n
X
(n,m)
λαj
(n,m)
+ βj
k+j=p
k≥0, j≥0
E (x; λ)
k
=2
k!
X
(n,m) x
βj
k
k!
k+j=p
k≥0, j≥0
,
and for p = m + n + r, r ≥ 1,
X
k+j=p
k≥0, j≥0
(n,m)
λαj
(n,m)
+ βj
E (x; λ)
k
=2
k!
X
k+j=p
k≥0, j≥0
b(p)
which proves the expression of A
n,m (x; λ).
(n,m) x
βj
k
k!
−λ
(n,m) Ek (x; λ)
X
γj
k+j=r−1
k≥0, j≥0
k!
,
Theorem 1 provides a recurrence relation of length max(n, m) from Bp−max(n,m) (x; λ) (resp.
Ep−max(n,m) (x; λ)) to Bp (x; λ) (resp. Ep (x; λ)) for p less than m + n. On the other hand, if p is
greater than m + n, we obtain also a recurrence relation for the same Apostol-type polynomials but with supplementary first terms from B0 (x; λ) (resp. E0 (x; λ)) to Bp−m−n−1 (x; λ) (resp.
Ep−m−n−1 (x; λ)).
Of course, Theorem 1 is valid for Bernoulli and Euler polynomials and also for Bernoulli and
Euler numbers which are particular Apostol-type polynomials. By this mean we recover known
results, see [1, 2, 22, 23, 24, 25, 26], as shown in the next sections.
7
4. Applications
In this section, we will consider the value of the parameter p with respect to n, m.
Let us consider the particular value p = m + n. For this particular case, from formulas (3.1)
and (3.3), we can prove the following corollary.
Corollary 1. For m ≥ 0, n ≥ 0,
λ
n X
n
k=0
λ
=
((m + n)x − n)xn−1 (x − 1)m−1 ,
m
X
n
m
k m
E
(x; λ) + (−1)
(−1)
En+k (x; λ)
k m+k
k
=
2xn (x − 1)m .
k
n X
k=0
m
X
m
Bn+k (x; λ)
k
Bm+k (x; λ) − (−1)m
(−1)k
k=0
k=0
(p)
(p)
bn,m
(x; λ) can be simplified as
Proof. For p = m + n, the expression of An,m (x; λ) and A
A(m+n)
(x; λ)
n,m
=λ
n X
n
k=0
k
m
Bm+k (x; λ) − (−1)
m
X
m
(−1)
Bn+k (x; λ)
k
k
k=0
and
b(m+n)
A
(x; λ) = λ
n,m
n X
n
k=0
k
Em+k (x; λ) + (−1)m
m
X
(−1)k
k=0
m
En+k (x; λ).
k
Using the identities
m X
m
j=0
j
m X
m
(−1)j j xj+1 = −m(1 − x)m−1 x2 ,
j=0
j
(−1)j xm−j = (x − 1)m
and formulas (3.1) and (3.3), we get Corollary 1. Let s, r ≥ 1. Setting p = m + n − s and p = m + n + r in Theorem 1, we will get Corollaries 2
and 3.
Corollary
2. For n ∈ N, m ∈ N, s ∈ N such that 1 ≤ s ≤ n+ m,
Pn
Pm
n
m
(m−s+k+1)s Bm−s+k (x; λ)−(−1)m k=0(−1)k
(n−s+k+1)s Bn−s+k (x; λ) =
λ k=0
k
k
Pm+n−s−1 m
(m + n − s − j)s+1 (−1)j xm+n−s−1−j ,
j=0
j
Pn
Pm
n
m
k m
λ k=0
(m−s+k+1)s Em−s+k (x; λ)+(−1)
(n − s + k + 1)s En−s+k (x; λ) =
k=0 (−1)
k
k
2
Pm+n−s m
(m + n − s + 1 − j)s (−1)j xm+n−s−j .
j=0
j
8
Remark 1. For λ = 1 and s = 1, Corollary 1 and Corollary 2 reduce to the following formulas
n m
X
X
n
m
Bm+k (x) − (−1)m
(−1)k
Bn+k (x) = ((m + n)x − n)xn−1 (x − 1)m−1 ,
k
k
k=0
k=0
n m
X
X
n
m
k m
Em+k (x) + (−1)
(−1)
En+k (x) = 2xn (x − 1)m ,
k
k
k=0
k=0
n X
n
k=0
m
X
k m
(m + k)Bm+k−1 (x)−(−1)
(−1)
(n + k)Bn+k−1 (x) =
k
k
m
k=0
m+n−2
X j=0
n X
n
k=0
k
(m + k)Em+k−1 (x)+(−1)m
m
X
m
(m + n − 1 − j)(m + n − j)(−1)j xm+n−2−j ,
j
(−1)k
k=0
m
(n + k)En+k−1 (x) =
k
2
m+n−1
X j=0
m
(m + n − j)(−1)j xm+n−1−j ,
j
proved by Wu and all [26] (generalizing Kaneko’s formula [22]) and by Momiyama in [25] for
Bernoulli numbers but without reference to a previous more general result by Agoh in [1].
Remark 2. For λ = 1, s = 3, m = n, x = 0 we find the formula proved by Chen and Sun in [2]
for the particular case of Bernoulli numbers.
Corollary
3. For n ∈ N, m ∈ N, r ∈ N, r ≥ 1, Pm
Pn
n Bm+r+k (x;λ)
m
k m Bn+r+k (x;λ)
λ k=0
k=0 (−1)
(m+1+k)r − (−1)
(n+1+k)r =
k
k
Pr−1 m+r−1−k
Pm m
m+n+r
n! m!
m
j xm+n+r−1−j
Bk (x; λ),
k=0
j=0 j (−1) (m+n+1−j)r−1 − λ(−1)
(n+m+r)!
m
k
Pm
n Em+r+k (x;λ)
m
k m En+r+k (x;λ)
λ k=0
k=0 (−1)
(m+1+k)r + (−1)
(n+1+k)r =
k
k
Pm m
Pr−1 m+r−1−k
m+n+r
n! m!
j xn+m+r−j
m
Ek (x; λ).
2 j=0
(−1) (m+n+1−j)r − λ(−1) (n+m+r)! k=0
j
m
k
Pn
Remark 3. When λ = 1, x = 0 and r = 1, the first formula of Corollary 3 reduces to a result due
to Gelfand [24].
5. Extension of Theorem 1
The formulas in Theorem 1 involve the Apostol-Bernoulli polynomials Bk (x; λ) or ApostolEuler polynomials Ek (x; λ). In the present section, using the same method of Padé approximants
as in Section 3, we prove the same type of recurrence relations for the sequences ( l−k Bk (x; λ))k
and ( l−k Ek (x; λ))k where l is some positive integer.
Theorem 2. For Apostol-Bernoulli polynomials, for n ∈ N, m ∈ N, p ∈ N, l ∈ N, if p ≤ m + n,
then
max(n,m)
l−1
P
P k p−1
P m (−1)j (n+m−j)! (x+k)p−1−j
n
m
(n+m−j)! Bp−j (x;λ)
λl
−(−1)j
=
λ
p−j
(p−j)!
l
lp−j
(p−1−j)! ,
j
j
j=0
j=0 j
k=0
9
and if p ≥ m + n + 1, then
max(n,m)
P
j=0
m
l−1
j
P kP
m
(x+k)p−1−j
(n+m−j)! Bp−j
l n
j m
λ
(n + m − j)! (−1)
λ
−(−1)
(p−j)! lp−j (x; λ) =
lp−j
(p−1−j)! −
j
j
j
j=0
k=0
p−m−n−1
P
p−1−n−k
p Bk (x;λ)
λl (−1)m n!m!
.
p!
lk
m
k
k=0
For Apostol-Euler polynomials, for n ∈ N, m ∈ N, p ∈ N, l ∈ N, if p ≤ m + n, then
max(n,m)
P
j=0
p
l−1
j
P
P
m
(n+m−j)! Ep−j (x;λ)
l n
l+j m
l−1
k
λ
−(−1)
= 2(−1)
(−λ)
(n+m−j)!(−1)
(p−j)!
lp−j
lp−j
j
j
j
j=0
k=0
(x+k)p−j
(p−j)! ,
and if p ≥ m + n + 1, then
max(n,m)
P
j=0
(n+m−j)!
l n
l+j m
λ
−(−1)
(p−j)!
j
j
j
m
(x+k)p−j
= 2 (−1)
(−λ)
(n+m−j)!(−1)
lp−j (p−j)! −
j
k=0
j=0
P
p−1−n−k
p Ek (x;λ)
p−m−n−1
l
m n! m!
.
λ (−1)
k=0
p!
lk
m
k
Ep−j (x;λ)
lp−j
l−1
l−1
P
k
m
P
Proof. The details of the proof will be omitted. For the Apostol-Bernoulli polynomials, multiply
the generating function (1.3) by (λl el t − 1) to obtain
t ex t (λl−1 e(l−1) t + · · · + λ et + 1) = (λl el t − 1)
∞
X
Bk (x; λ)
k=0
tk
,
k!
(5.1)
and substitute el t in the right hand-side member of (5.1) by its Padé approximant [n, m](lt)
as in the proof of Theorem 1. For the Apostol-Euler polynomials, we must multiply (1.5) by
(λl el t + (−1)l−1 ) to obtain
xt
2 e (λ
l−1 (l−1) t
e
−λ
l−2 (l−2) t
e
l−1
+ · · · + (−1)
l lt
) = (λ e + (−1)
l−1
)
∞
X
k=0
Ek (x; λ)
tk
.
k!
Remark 4. If λ = 1, x = 0 and p = m + n − 1 the first formula of Theorem 2 reduces to a result
due to Gessel [23]. If λ = 1, x = 0 and p = m + n − 3 it reduces to a result proved in [2].
6. Limit case: m=ρ n, n → ∞.
(p)
Let us recall the expression for An,m :
m
n X
X
n (n + m − j)!
j m (n + m − j)!
(p)
An,m (x; λ) = λ
Bp−j (x; λ) −
(−1)
Bp−j (x; λ). (6.1)
j
j
(p − j)!
(p − j)!
j=0
j=0
If n, m tend to infinity, then of course, p ≤ m + n and so by Theorem 1
A(p)
n,m (x; λ)
=
p−1 X
m
j=0
j
(n + m − j)!(−1)j
xp−1−j
.
(p − 1 − j)!
Let us assume that m = ρ n with n going to infinity. After dividing the previous identity by
(n + m − p)!, we get
p−1 (p)
X
An,ρ n (x; λ)
xp−1−j
ρ n (n + ρ n − j)!
=
(−1)j
.
(n + ρ n − p)! j=0 j
(n + ρ n − p)!
(p − 1 − j)!
10
(6.2)
Using
nj
n
∼
, (j ≥ 0)
j n→∞ j!
and
n+ρn−j
n+ρn−p
∼
n→∞
(1 + ρ)p−j p−j
n , (0 ≤ j ≤ p)
(p − j)!
we get from (6.1)
(p)
An,ρ n (x; λ)
(n + ρ n − p)!
∼
n→∞
p np X p
(λ − (−ρ)j )(1 + ρ)p−j Bp−j (x; λ).
p! j=0 j
Similarly, the right hand-side member of Eq. (6.2) satisfies
p−1 X
ρ n (n + ρ n − j)!(−1)j xp−1−j
j=0
j
(n + ρ n − p)!(p − 1 − j)!
p−1
∼
np X j
ρ (1 + ρ)p−j (−1)j
(p − 1)! j=0
∼
np
(1 + ρ)(x (1 + ρ) − ρ)p−1 .
(p − 1)!
n→∞
n→∞
p − 1 p−1−j
x
j
Then, we are led to the following theorem.
Theorem 3. For all complex number ρ and for all integer p, the following formulas hold:
p X
p
j=0
j
p X
p
j=0
j
(λ − (−ρ)j )(1 + ρ)p−j Bp−j (x; λ)
=
p (1 + ρ) (x (1 + ρ) − ρ)p−1 ,
(6.3)
(λ + (−ρ)j )(1 + ρ)p−j Ep−j (x; λ)
=
2 (x (1 + ρ) − ρ)p .
(6.4)
Proof. It can be found that the generating functions of both sides of (6.3) are (1 + ρ)te(x(1+ρ)−ρ)t
and the generating functions of both sides of (6.4) are 2e(x(1+ρ)−ρ)t .
Remark 5. Set X = x − ρ/(1 + ρ), the previous identities turn to
λBp (X + 1; λ) − Bp (X; λ) = p X p−1 , (p ≥ 0),
and
λEp (X + 1; λ) + Ep (X; λ) = 2 X p , (p ≥ 0)
which can be found in [3, 9].
7. Lacunary recurrence relations
Typically, the calculation of Apostol-type polynomials requires the calculation of all the previous polynomials. For the particular case of Bernoulli numbers, in [27], Ramanujan proved a
recurrence relation between the Bernoulli numbers with a gap of length 6. Lehmer [28] in 1935
extended these methods to Euler numbers, Genocchi numbers and Lucas numbers. In [29], using
the ”multisectioning” technique, Hare constructed lacunary recursion formulas of arbitrary size.
In this section, we use Theorem 3 to get recurrence relations for the sequences (B4k+a (x))k ,
(E4k+a (x))k , with a ∈ {0, 1, 2, 3}. For the particular case x = 0, we will give relations for ( B6k+a )k ,
(E6k+a )k , where a ∈ {0, 2, 4}.
11
The sequence of Bernoulli polynomials (Bk (x) := Bk (x; 1))k satisfies the property (see Theorem 3) p−1
Pp
p
ρ
j
−j
x − 1+ρ
, for each complex number ρ.
j=0 j (1 − (−ρ) )(1 + ρ) Bp−j (x) = p
2
If we take
ρ = i, (i = −1), then
Pp
Pp
p
p −k/2 −i k π/4
j
−j
e
− (−1)k ei k π/4 Bp−k (x),
j=0 j (1 − (−ρ) )(1 + ρ) Bp−j (x) =
k=0 k 2
and
Pp−1 p − 1
p−1−k
p−1
p−1
1−p
1−p
(2x − 1)
(−i)k .
p (x − ρ/(1 + ρ))
= p2
(2x − 1 − i)
= p2
k=0
k
Thus
p X
p
k
k=0
2
−k/2
−i k π/4
e
k i k π/4
− (−1) e
Bp−k (x) = p 2
1−p
p−1 X
p−1
k=0
k
p−1−k
(2x − 1)
(−i)k .
(7.1)
Let us calculate the coefficient ck = e−i k π/4 − (−1)k e
k
≡
0(4)
k
≡
1(4)
k
≡
2(4)
k
≡
3(4)
i k π/4
:
ck = 0,
√
ck = (−1)(k−1)/4 2,
ck = −2(−1)(k−2)/4 i,
√
ck = −(−1)(k−3)/4 2.
So, if we consider only the imaginary part of formula (7.1), this will provide a recurrence relation
with a gap of length 4. After simplification the following formula
p
p−1 X
X
p −k/2
p−1
p−1−k
(k−2)/4
−p
2
(−1)
Bp−k (x) = p 2
(2x − 1)
(−1)(k−1)/2
k
k
k=1
k odd
k≡2(4)
holds. So, according to the value of p modulo 4, for all integer q ≥ 1
q X
4q
(−4)j B4j−2 (x)
4j − 2
=
q
1−2q
(−1) 4 q 2
j=1
q X
4q + 1
(−4)j B4j−1 (x)
4j − 1
j=0
4j
q X
4q + 3
j=0
4j + 1
(−4)j B4j (x)
k=1
k odd
=
−2q
q
(−1) (4q + 1) 2
j=1
q X
4q + 2
4q−1
X
k−1
4q − 1
4q−1−k
(2x − 1)
(−1) 2 ,
k
4q X
k−1
4q
4q−k
(2x − 1)
(−1) 2 ,
k
k=1
k odd
=
(−1)q (4q + 2) 2−1−2q
4q+1
X
k−1
4q + 1
4q+1−k
(2x − 1)
(−1) 2 ,
k
k−1
4q + 2
4q+2−k
(2x − 1)
(−1) 2 .
k
k=1
k odd
(−4)j B4j+1 (x)
=
(−1)q (4q + 3) 2−2−2q
4q+2
X
k=1
k odd
The sequence of Euler polynomials (Ek (x) := Ek (x; 1))k satisfies the property (see Theorem 3)
p
p
j
−j
j=0 j (1 + (−ρ) )(1 + ρ) Ep−j (x) = 2 (x − ρ/(1 + ρ)) , for each complex number ρ.
If we takeρ = i, (i2 = −1), then
Pp
Pp
p
p −k/2 −i k π/4
j
−j
(1
+
(−ρ)
)(1
+
ρ)
E
(x)
=
e
+ (−1)k ei k π/4 Ep−k (x),
p−j
j=0 j
k=0 k 2
Pp
12
p
p
and 2 (x − ρ/(1 + ρ)) = 21−p (2x − 1 − i) = 21−p
p
p−k
(−i)k .
k=0 k (2x − 1)
Pp
Thus
p X
p
k=0
k
2
−k/2
e
−i k π/4
k i k π/4
+ (−1) e
Ep−k (x) = 2
p X
p
1−p
k=0
k
p−k
(2x − 1)
(−i)k .
(7.2)
Let us calculate the coefficient ĉk = e−i k π/4 + (−1)k ei k π/4 :
k
≡ 0(4)
k
≡ 1(4)
k
≡ 2(4)
k
≡ 3(4)
ĉk = 2 (−1)k/4 ,
√
ĉk = i 2(−1)(k+3)/4 ,
ĉk = 0,
√
ĉk = i 2(−1)(k+1)/4 .
So, if we consider only the real part of formula (7.2), this will provide recurrence with a gap of
length 4. After simplification the following formula
p
p
X
X
p −k/2
p
p−k
k/4
−p
2
(−1) Ep−k (x) = 2
(2x − 1)
(−1)k/2
k
k
k=0
k even
k≡0(4)
holds. Thus
q X
4q
4j
j=0
q X
4q + 1
j=0
4j + 1
4j + 2
(−1)q 2−2q
4q
X
4q
4q−k
(2x − 1)
(−1)k/2 ,
k
k=0
k even
j
(−4) E4j+1 (x)
=
q
−1−2q
(−1) 2
4q+1
X
(−4)j E4j+2 (x)
=
(−1)q 2−2−2q
4q+2
X
4j + 3
j
(−4) E4j+3 (x)
=
q
−3−2q
(−1) 2
4q+3
X
4q + 1
4q+1−k
(2x − 1)
(−1)k/2 ,
k
k=0 ,
k even
q X
4q + 3
j=0
=
k=0
k even
q X
4q + 2
j=0
(−4)j E4j (x)
k=0
k even
4q + 2
4q+2−k
(2x − 1)
(−1)k/2 ,
k
4q + 3
4q+3−k
(2x − 1)
(−1)k/2 .
k
In Theorem 3, if we set λ = 1, x = 0 in formula (6.3) and λ = 1, x = 1/2 in formula (6.4) then
p X
p
j=0
p X
p
j=0
j
j
(1 − (−ρ)j )(1 + ρ)p−j Bp−j
(1 + (−ρ)j )
1+ρ
2
=
p (1 + ρ)(−ρ)p−1 ,
=
2
p−j
Ep−j
1−ρ
2
(7.3)
p
.
(7.4)
√
Setting ρ = µ = −1/2 + i 3/2 in (7.3) and (7.4), we can derive some recurrence relations with
gap of length 6, for Bernoulli and Euler numbers (the first three identities have been proved by
Ramanujan in [27]). See also [28, 30].
13
Theorem 4. For odd integer p ≥ 3, we have
p−1 X
p
p
Bk = −
if p
k
6
≡
1(6),
(7.5)
p−1 X
p
p
Bk =
if p
k
3
≡
3(6),
(7.6)
p−1 X
p
p
if p
Bk =
k
3
≡
5(6).
(7.7)
k=0
k≡4(6)
k=0
k≡0(6)
k=0
k≡2(6)
For even integer p, we have
p
X
p −k
2 Ek = 21−p 1 + (−3)p/2
if p
k
≡ 0(6),
(7.8)
p
X
p −k
Ep + 3
2 Ek = 21−p 1 + (−3)p/2
if p
k
≡ 2(6),
(7.9)
p
X
p −k
2 Ek = 21−p 1 + (−3)p/2
if p
k
≡ 4(6).
(7.10)
2−p Ep + 3
k=0
k≡0(6)
2
−p
k=0
k≡2(6)
2−p Ep + 3
k=0
k≡4(6)
Proof. Let us assume that p is an odd integer greater than 3, then from (7.3), we obtain
p
X
j=1
p−j≡0(2)
p
(1 − (−ρ)j )(1 + ρ)−j Bp−j
j
=
p
(1 + ρ)1−p (1 + (−ρ)p−1 ).
2
(7.11)
√
Let us take ρ = µ, with µ = −1/2 + i 3/2. For odd integer p ≥ 3, using 1/(1 + µ) = −µ, we get
j !
p−1 !
p
X
−µ
p
µ
p
(1 + µ)−j −
Bp−j =
(1 + µ)1−p +
,
j
1
+
µ
2
1+µ
j=1
j≡1(2)
then
p
X
p
(−µ)j − µ2j Bp−j
j
j=1
j≡1(2)
=
p p−1
µ
+ µ2p−2 .
2
(7.12)
The coefficient dj = (−µ)j − µ2j have some repeated values according to the value modulo 6 of
the odd integer j:
j ≡ ±1(6),
dj = 1
j ≡ 3(6),
dj = −2.
Then (7.12) indicates
p
X
j=1
p−j≡±1(6)
p
Bp−j − 2
j
p
X
j=1
p−j≡3(6)
p p−1
p
µ
+ µ2p−2 .
Bp−j =
j
2
14
By difference with the relation
p
X
j=1
p−j≡0(2)
p
p
Bp−j =
j
2
obtained from (7.11) by setting ρ = 0, it arises
3
p−1
X
j=0
p−j≡3(6)
p
p
p
Bj = − µp−1 + µ2p−2 + ,
j
2
2
which proves formulas (7.5), (7.6) and (7.7).
In a similar way, by making use of (7.4), we obtain formulas (7.8) – (7.10). Acknowledgment. I would like to thank the anonymous referees for pointing out several
errors in earlier versions of this paper.
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